$\sup(A\cdot B) = (\sup A)(\sup B)$ where $A$ and $B$ bounded sets of positive real numbers

Let $L:=(\sup A)(\sup B)$. By definition of "supremum", one needs to show the following two things,

• $L$ is an upper bound for $A\cdot B$;

• $L$ is the smallest upper bound for $A\cdot B$.

The first statement is very easy to prove: since for every $a\in A$ and every $b\in B$, one has $a\le \sup A$ and $b\le \sup B$, which immediately implies that $ab\le (\sup A)(\sup B)$.

To prove the second statement, assume that $L'$ is an upper bound of $A\cdot B$. We want to show that $L\le L'$. But the assumption of $L'$, we have for every $a\in A$,

$$\text{for every } b\in B: ab\le L'$$

which is equivalent to

$$\text{for every } b\in B: b\le \frac{1}{a}\cdot L'$$

So it follows that $\sup B\le L'/a$. Thus for every $a\in A$, one has

$$ a\le \frac{L'}{\sup B} $$

and thus $\sup(A)\le \frac{L'}{\sup B}$, which implies that $L\le L'$.

Another way to write the proof is to use an alternative equivalent definition of supremum:

For a bounded set of real numbers $A$, $L=\sup A$ if

• $L$ is an upper bound for $A$, i.e., for all $a\in A$: $a\le L$;
• any number smaller than $L$ is not an upper bound of $A$: for every $\epsilon>0$, there exists $a\in A$ such that $a>L-\epsilon$.

As in the other answer, we let $L=(\sup A)(\sup B)$ and we show that

• $L$ is an upper bound for $A\cdot B$;

• for every $\epsilon>0$, there exists $a\in A$ and $b\in B$ such that $$ ab> L-\epsilon\tag{1} $$

The first statement is proved in the same way as in the other answer. Let us show the second one. Let $\epsilon >0$. We want to find $a$ and $b$ so that (1) holds.

By the definition of supremum, for every $\epsilon'>0$, we can find $a\in A$ and $b\in B$ so that

$$ a+\epsilon'>\sup A,\quad b+\epsilon' >\sup B $$

which gives $$ab+(a+b)\epsilon'+(\epsilon')^2>L\tag{2}$$

So, if we pick $\epsilon'$ so that $$ \epsilon \ge (\sup(A)+\sup(B)+1)\epsilon' $$ and correspondingly, the $a$ and $b$ that (2) holds, we have $$ ab+\epsilon \ge ab +(\sup(A)+\sup(B)+1)\epsilon'\ge ab+(a+b+\epsilon')\epsilon'>L. $$ We are done.

To further emphasize the "fake difficulty", we can show that most of this proof is completely "formal", in that it holds for very general reasons that have little to do with real numbers. See the bottom for a detailed, elementary proof that explicitly illustrates that the proof is almost entirely a manipulation of definitions and basic logic.

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Let $\mathcal C$, $\mathcal D$, and $\mathcal E$ be arbitrary categories and $\odot : \mathcal D \times \mathcal E \to \mathcal C$ be a bifunctor. Write $\int^{I : \mathbb I} D(I)$ for the colimit of the diagram $D$ which is just a functor from some small category $\mathbb I$ to another category. If $\odot$ is cocontinuous in both arguments, then $$\left(\int^{I:\mathbb I}D_1(I)\right) \odot \left(\int^{J:\mathbb J}D_2(J)\right) \cong \int^{(I, J) : \mathbb I \times \mathbb J} D_1(I)\odot D_2(J)$$

All left adjoints are cocontinuous, so it would suffice to show that both $({-})\odot E$ and $D \odot ({-})$ are left adjoints for all $E \in \mathrm{Ob}(\mathcal E)$ and $D \in \mathrm{Ob}(\mathcal D)$.

All these facts are easily provable facts of basic category theory and coend calculus.

Now, let's instantiate it. Given any partially ordered set, $S$, we can make a category whose objects are the elements of $S$ and whose hom-sets have at most one arrow with $\mathrm{Hom}(s, s')$ being non-empty exactly when $s \leq s'$. A functor between such categories is exactly a monotonic function. The supremum of $A \subseteq S$ is the colimit of the diagram $\mathsf D A \to S$ where $\mathsf D$ turns a set into a discrete category, and we're identifying the partially ordered set with its corresponding category.

We'll choose $S = \mathbb R^{+}$, the set of positive reals with the usual ordering. We'll define $\odot : \mathbb R^+ \times \mathbb R^+ \to \mathbb R^+$ to be $x \odot y = xy$. To be a (bi)functor, it must be monotonic in each argument which is a basic fact of real number multiplication (for positive numbers). $\odot$ is obviously commutative, so we can complete our proof if we show that $x \odot ({-})$ is a left adjoint which it is via the right adjoint $({-})/x$. The adjunction condition is simply $$xy \leq z \iff y \leq z/x$$ for all $x$, $y$, and $z$ in $\mathbb R^+$ which is, again, a basic fact of (positive) real numbers.

Obviously it would be silly to go through these general results that have to track a bunch of detail that is degenerate in this example. My point is 1) the core pieces of the proof generalize massively if appropriately formulated, and 2) taking this abstract perspective and having a good handle on categorical intuitions can feed back into elegant proofs and a start to teaching categorical intuitions. In fact, a lot of the useful logical relations can be motivated by categorical intuitions. For example, below we'll use $(\exists x.P(x)) \to Q \iff \forall x. P(x) \to Q$ which is just continuity of right adjoints, and indiscernability of identicals is a special case of the Yoneda lemma.

Here's the proof I'd actually present in an undergrad class. (Well, I'd probably not present it in such detail.)

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First, I'd define the supremum (in general) via the following, for $A \subseteq S$ for a partially ordered set, $S$, $\sup A$ is the unique element of $S$ satisfying $$\sup A \leq z \iff \forall a \in A. a \leq z$$ for all $z \in S$ which is a natural definition (and secretly is also the representability formulation of coproducts/colimits).

As lemmas, we'd have $ab \leq z \iff a \leq z/b$ (exercise; this is the only vaguely $\mathbb R^+$-specific result here) and

$$\begin{flalign} \sup (A\cdot B) \leq z & \iff \forall x \in (A \cdot B). x \leq z \tag{definition of $\sup$} \\ & \iff \forall x. (x \in (A \cdot B)) \to x \leq z \tag{expand shorthand} \\ & \iff \forall x. (\exists a \in A. \exists b \in B. x = ab) \to x \leq z \tag{definition of $A \cdot B$} \\ & \iff \forall x. \forall a \in A. \forall b \in B. (x = ab) \to x \leq z \tag{$(\exists x.P(x)\to Q \iff \forall x. P(x) \to Q$} \\ & \iff \forall a \in A. \forall b \in B. \forall x. (x = ab) \to x \leq z \tag{commutativity of $\forall$}\\ & \iff \forall a \in A. \forall b \in B. ab \leq z \tag{indiscernibility of identicals} \end{flalign}$$

The proof is then:

$$\begin{flalign} (\sup A)(\sup B) \leq z & \iff \sup A \leq z/(\sup B) \tag{lemma} \\ & \iff \forall a \in A. a \leq z/(\sup B) \tag{definition of $\sup$}\\ & \iff \forall a \in A. a(\sup B) \leq z \tag{lemma} \\ & \iff \forall a \in A. (\sup B) \leq z/a \tag{commutativity, lemma} \\ & \iff \forall a \in A. \forall b \in B. b\leq z/a \tag{definition of $\sup$} \\ & \iff \forall a \in A. \forall b \in B. ab \leq z \tag{commutativity, lemma} \\ & \iff \sup (A\cdot B) \leq z \tag{other lemma} \end{flalign}$$

To get $(\sup A)(\sup B) = \sup (A\cdot B)$ simply choose $z$ to be $(\sup A)(\sup B)$ and $\sup (A\cdot B)$.